一道三角函数题...

-sinC=sinA+sinB
-cosC=cosA+cosB
(sinC)^2+(cosC)^2=(sinA+sinB)^2+(cosA+cosB)^2
所以，
1=2+2(sinAsinB+cosAcosB)
即
cos(A-B)=sinAsinB+cosAcosB
=-1/2

sina+sinb=-sinc ,两边平方(sina+sinb)＾2=(-sinc)＾2
sin＾2a+2sina*sinb+sin＾2b=sin＾2c
cosa+cosb=-cosc,两边平方(cosa+cosb)＾2=(-cosc)＾2,
cos＾2a+2cosa*cosb+cos＾2b=cos＾2c
(sin＾2a+2sina*sinb+sin＾2b)+
(cos＾2a+2cosa*cosb+cos＾2b)=sin＾2c+cos＾2c,
2+2cos(a-b)=1,
cos(a-b) =-1/2